Clinical pharmacokinetics worksheet with answers

Free printable Clinical pharmacokinetics worksheet: 6 problems from easy to exam level, with an answer key. Print it or save it as a PDF.

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1.Easy
0.1110t (h)C (mg/L)8 mg/L7.5 hC = ?

Practice exercise (made-up values, not clinical guidance). Just after an IV dose, the tobramycin level is 8 mg/L. The half-life is 2.5 h, and the drug is eliminated by first-order kinetics. What is the level C 7.5 h later?

C₀ = 8 mg/Lt½ = 2.5 ht = 7.5 h
C = mg/L
2.Easy
Vd = 0.7 L/kgC = 30 mg/LLD = ?80 kg

Practice exercise (made-up values, not clinical guidance). A patient who weighs 80 kg needs a loading dose of vancomycin to reach a level of 30 mg/L straight away. The volume of distribution is 0.7 L/kg. Find the loading dose LD in mg.

80 kgVd = 0.7 L/kgC = 30 mg/L
LD = mg
3.Medium
VCss​ = 10 mg/LR = ? mg/hCL = 0.035 L/kg/h92 kg

Practice exercise (made-up values, not clinical guidance). A 92 kg patient receives a theophylline infusion. The clearance is 0.035 L/kg/h. What infusion rate R in mg/h gives a steady-state level of 10 mg/L?

92 kgCL = 0.035 L/kg/hCss = 10 mg/L
R = mg/h
4.Medium
110t (h)C (mg/L)2 h6 mg/L6 h2.4 mg/L

Practice exercise (made-up values, not clinical guidance). After an IV dose of tobramycin, the level is 6 mg/L at 2 h and 2.4 mg/L at 6 h. Assume first-order elimination. Find the elimination rate constant k.

C1 = 6 mg/L (2 h)C2 = 2.4 mg/L (6 h)
k = h⁻¹
5.Hard
Vd = 0.7 L/kg10 → 20 mg/LLD = ? mg85 kg · S = 0.92 · F = 1

Practice exercise (made-up values, not clinical guidance). An 85 kg patient has a phenytoin level of 10 mg/L; the target is 20 mg/L. The drug is given intravenously as phenytoin sodium (S = 0.92), and the volume of distribution is 0.7 L/kg. What extra loading dose LD raises the level to the target?

85 kgVd = 0.7 L/kgC = 10 → 20 mg/LS = 0.92
LD = mg
6.Hard
VCss,avg​ = 12 mg/LD = ? mg · q12hCL = 0.04 L/kg/h80 kg · S = 0.8 · F = 1

Practice exercise (made-up values, not clinical guidance). An 80 kg patient takes oral aminophylline (S = 0.8, F = 1) every 12 h. The theophylline clearance is 0.04 L/kg/h. What dose D in mg gives an average steady-state level of 12 mg/L?

CL = 0.04 L/kg/hτ = 12 hS = 0.8F = 180 kgCss,avg = 12 mg/L
D = mg

Answer key

  1. 1. C = 1 mg/L
  2. 2. LD = 1680 mg
  3. 3. R = 32.2 mg/h
  4. 4. k = 0.2291 h⁻¹
  5. 5. LD = 646.7 mg
  6. 6. D = 576 mg

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